2009年广东省中山市初中毕业生学业考试

2009年广东省中山市初中毕业生学业考试
数 学
说明:
1.全卷共4页，考试用时100分钟，满分为120分.
2.答卷前，考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号、姓名、试室号、座位号．用2B 铅笔把对应该号码的标号涂黑．
3.选择题每小题选出答案后，用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑，如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试题上．
4．非选择题必须用黑色字迹钢笔或签字笔作答、答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液．不按以上要求作答的答案无效．
5．考生务必保持答题卡的整洁．考试结束时，将试卷和答题卡一并交回．
一、选择题（本大题5小题，每小题3分，共15分）在每小题列出的四个选项中，只有一个是正确的，请把答题卡上对应题目所选的选项涂黑． 1．4的算术平方根是（ ） A ．2±
B ．2
C
．
D
2．计算32
()a 结果是（ ） A ．6
a
B ．9
a
C ．5
a
D ．8
a
3．如图所示几何体的主（正）视图是（ ）
C ．
4．《广东省2009年重点建设项目计划（草案）》显示，港珠澳大桥工程估算总投资726亿元，用科学记数法表示正确的是（ ）
A ． 10
7.2610⨯元 B ．9
72.610⨯元 C ．11
0.72610⨯元 D ．11
7.2610⨯元
5．方程组22
30
10x y x y +=⎧⎨+=⎩
的解是（ ） A ．1113x y =⎧⎨
=⎩
22
13x y =-⎧⎨
=-⎩ B ．12123
3
11x x y y ==-⎧⎧⎨⎨
=-=⎩⎩ C ． 12123
31
1x x y y ==-⎧⎧⎨
⎨==-⎩⎩ Ｄ．1212
1
1
3
3x x y y ==-⎧⎧⎨⎨=-=⎩⎩ 二、填空题：（本大题5小题，每小题4分，共20分）请将下列各题的正确答案填写在答题卡相应的位置上．
6．分解因式2
2
33x y x y --- ．
7．已知O ⊙的直径8cm AB C =，为O ⊙上的一点，30BAC ∠=°，则BC = cm ．
8．一种商品原价120元，按八折（即原价的80%）出售，则现售价应为 元．
9．在一个不透明的布袋中装有2个白球和n 个黄球，它们除颜色不同外，其余均相同．若从中随机摸出一个球，摸到黄球的概率是
45
，则n =_____________．
10．用同样规格的黑白两种颜色的正方形瓷砖，按下图的方式铺地板，则第（3）个图形中有黑色瓷砖 块，第n 个图形中需要黑色瓷砖________块（用含n 的代数式表示）．
……
（1） （2） （3）
三、解答题（一）（本大题5小题，每小题6分，共30分） 11．（本题满分6分）计算：1
sin 30π+32
-
+0°+()． 12．（本题满分6分）解方程2
21
11
x x =--- 13．（本题满分6分）如图所示，ABC △是等边三角形， D 点是AC 的中点，延长BC 到E ，使CE CD =，
（1）用尺规作图的方法，过D 点作DM BE ⊥，垂足是M （不写
作法，保留作图痕迹）； （2）求证：BM EM =．
14．（本题满分6分）已知：关于x 的方程2
210x kx +-=
（1）求证：方程有两个不相等的实数根；
（2）若方程的一个根是1-，求另一个根及k 值．
15．（本题满分6分）如图所示，A 、B 两城市相距100km ，现
计划在这两座城市间修建一条高速公路（即线段AB ），经测量，
森林保护中心P 在A 城市的北偏东30°和B 城市的北偏西45°的方向上，已知森林保护区的范围在以P 点为圆心，50km 为半径的圆形区域内，请问计划修建的这条高速公路会不会穿越保护
1.732 1.414）
第7题图
B
第10题图 A
D
第13题图
30° A B
F
E P
45°
第15题图
四、解答题（二）（本大题4小题，每小题7分，共28分） 16．（本题满分7分）某种电脑病毒传播非常快，如果一台电脑被感染，经过两轮感染后就会有81台电脑被感染．请你用学过的知识分析，每轮感染中平均一台电脑会感染几台电脑？若病毒得不到有效控制，3轮感染后，被感染的电脑会不会超过700台？ 17．（本题满分7分）某中学学生会为了解该校学生喜欢球类活动的情况，采取抽样调查的方法，从足球、乒乓球、篮球、排球等四个方面调查了若干名学生的兴趣爱好，并将调查的结果绘制成如下的两幅不完整的统计图（如图1，图2要求每位同学只能选择一种自己喜欢的球类；图中用乒乓球、足球、排球、篮球代表喜欢这四种球类中的某一种球类的学生人数），请你根据图中提供的信息解答下列问题：
（1）在这次研究中，一共调查了多少名学生？
（2）喜欢排球的人数在扇形统计图中所占的圆心角是多少度？ （3）补全频数分布折线统计图．
18．（本题满分7分）在ABCD 中，10AB =，AD m =，60D ∠=°，以AB 为直径作O ⊙， （1）求圆心O 到CD 的距离（用含m 的代数式来表示）； （2）当m 取何值时，CD 与O ⊙相切．
19．（本题满分7分）如图所示，在矩形ABCD 中，12AB AC =，=20，两条对角线相交于点O ．以OB 、OC 为邻边作第1个平行四边形1OBB C ，对角线相交于点1A ，再以11A B 、
1A C 为邻边作第2个平行四边形111A B C C ，对角线相交于点1O ；再以11O B 、11O C 为邻边
图2
乒乓球
20% 足球
排球 篮球
40%
图1 第17题图 第18题图
作第3个平行四边形1121O B B C ……依次类推． （1）求矩形ABCD 的面积；
（2）求第1个平行四边形1OBB C 、第2个平行四边形111A B C C 和第6个平行四边形的面积．
五、解答题（三）（本大题3小题，每小题9分，共27分） 20、（本题满分9分）
（1）如图1，圆心接ABC △中，AB BC CA ==，OD 、OE 为O ⊙的半径，OD BC ⊥于点F ，OE AC ⊥于点G ，
求证：阴影部分四边形OFCG 的面积是ABC △的面
积的13．
（2）如图2，若DOE ∠保持120°角度不变， 求证：当DOE ∠绕着O 点旋转时，由两条半径和
ABC △的两条边围成的图形（图中阴影部分）面积
始终是ABC △的面积的1
3
．
21．（本题满分9
分）小明用下面的方法求出方程30=的解，请你仿照他的方法求出下面另外两个方程的解，并把你的解答过程填写在下面的表格中．
A 1
O 1
A 2
B 2 B 1
C 1 B C 2
A O
D
第19题图 C 第20题图
D 图1 图2
22．（本题满分9分）正方形ABCD 边长为4，M 、N 分别是BC 、CD 上的两个动点，当M 点在BC 上运动时，保持AM 和MN 垂直，
（1）证明：Rt Rt ABM MCN △∽△；
（2）设BM x =，梯形ABCN 的面积为y ，求y 与x 之间的函数关系式；当M 点运动到什么位置时，四边形ABCN 面积最大，并求出最大面积； （3）当M 点运动到什么位置时Rt Rt ABM AMN △∽△，求x 的
值．
广东省中山市2009年初中毕业生学业考试
数学试题参考答案及评分建议
一、选择题（本大题5小题，每小题3分，共15分） 1．B 2．A 3．B 4．A 5．D
二、填空题（本大题5小题，每小题4分，共20分）
6．()(3)x y x y +-- 7．4 8．96 9．8 10．10，31n + 三、解答题（一）（本大题5小题，每题6分，共30分） 11．解：原式=
11
3122
+-+ ·
·················································································· 4分 =4． ······························································································· 6分
12．解：方程两边同时乘以(1)(1)x x +-， ······························································· 2分
2(1)x =-+， ···································································································· 4分
3x =-， ·
·········································································································· 5分 经检验：3x =-是方程的解． ················································································ 6分 13．解：（1）作图见答案13题图，
··························································· 2分 （2）ABC △是等边三角形，D 是AC 的中点，
BD ∴平分ABC ∠（三线合一）
， 2ABC DBE ∴∠=∠． ·
························································································ 4分 N
D
A C B
M
第22题图
答案13题图
A
C B
D
E M
CE CD =，
CED CDE ∴∠=∠．
又ACB CED CDE ∠=∠+∠，
2ACB E ∴∠=∠． ·
···························································································· 5分 又ABC ACB ∠=∠， 22DBC E ∴∠=∠， DBC E ∴∠=∠， BD DE ∴=． 又DM BE ⊥，
BM EM ∴=． ·
································································································· 6分 14．解：（1）2
210x kx +-=，
2242(1)8k k ∆=-⨯⨯-=+， ·············································································· 2分
无论k 取何值，2
k ≥0，所以2
80k +>，即0∆>，
∴方程2210x kx +-=有两个不相等的实数根． ························································ 3分
（2）设2
210x kx +-=的另一个根为x ，
则12k x -=-
，1(1)2x -=-，·············································································· 4分 解得：1
2
x =，1k =，
∴2210x kx +-=的另一个根为1
2
，k 的值为1． ····················································· 6分
15．解：过点P 作PC AB ⊥，C 是垂足，
则30APC ∠=°，45BPC ∠=°， ····································· 2分
tan30AC PC =°，tan 45BC PC =°，
AC BC AB +=， ·
······················································· 4分 tan30tan 45100PC PC ∴+
=°°，
11003PC ⎛⎫∴+= ⎪ ⎪⎝⎭， ···················································
5分 50(350(3 1.732)63.450PC ∴=⨯->≈≈，
答：森林保护区的中心与直线AB 的距离大于保护区的半径，所以计划修筑的这条高速公路不会穿越保护区．································································································ 6分 四、解答题（二）（本大题4小题，每小题7分，共28分） 16．解：设每轮感染中平均每一台电脑会感染x 台电脑， ············································ 1分 依题意得：1(1)81x x x +++=， ··········································································· 3分
2(1)81x +=，
答案15题图
A B
F E P C
19x +=或19x +=-，
12810x x ==-，（舍去），··················································································· 5分 33(1)(18)729700x +=+=>． ············································································ 6分
答：每轮感染中平均每一台电脑会感染8台电脑，3轮感染后，被感染的电脑会超过700台． ························································································································ 7分 17．解：（1）2020%100÷=（人）． ····································································· 1分
（2）30
100%30%100
⨯=， ·
·················································································· 2分 120%40%30%10%---=，
36010%36⨯=°°． ·
··························································································· 3分 （3）喜欢篮球的人数：40%10040⨯=（人）， ························································ 4分 喜欢排球的人数：10%10010⨯=（人）． ································································ 5分
······················· 7分
18．解：（1）分别过A O ，两点作AE CD OF CD ⊥⊥，，垂足分别为点E ，点F ，
AE OF OF ∴∥，就是圆心O 到CD 的距离． 四边形ABCD 是平行四边形，
AB CD AE OF ∴∴=∥，． ·
················································································· 2分
在Rt ADE △中，60sin sin 60AE AE
D D AD AD
∠=∠==°
，，°，
222
AE AE m OF AE m m ====，，， ························································ 4分 答案17题图
答案18
题图（1）
答案18题图（2）
圆心到CD 的距离OF
． ··········································································· 5分 （2）
3
OF =
， 为O ⊙的直径，且10AB =，
当5OF =时，CD 与O ⊙相切于F 点，
5m ==， ··················································································· 6分
当3
m =
时，CD 与O ⊙相切． ······································································· 7分 19．解：（1）在Rt ABC △
中，
16BC =，
1216192ABCD S AB BC ==⨯=矩形． ······································································ 2分
（2）
矩形ABCD ，对角线相交于点O ，
4ABCD OBC S S ∴=△． ···························································································· 3分
四边形1OBB C 是平行四边形，
11OB CB OC BB ∴∥，∥，
11OBC B CB OCB B BC ∴∠=∠∠=∠，．
又
BC CB =，
1OBC B CB ∴△≌△，
11
2962OBB C OBC ABCD S S S ∴==
=△， ·
······································································ 5分 同理，1111111
48222
A B C C OBB C ABCD S S S ==⨯⨯=， ························································ 6分
第6个平行四边形的面积为61
32
ABCD S =． ······························································· 7分
五、解答题（三）（本大题3小题，每小题9分，共27分） 20．证明：（1）如图1，连结OA OC ，， 因为点O 是等边三角形ABC 的外心，
所以Rt Rt Rt OFC OGC OGA △≌△≌△． ····························· 2分
2OFCG OFC OAC S S S ==△△，
答案20题图（1）
A
E O G F
B
C
D
因为1
3OAC ABC S S =△△， 所以1
3
OFCG
ABC S S =△． ·
······················································································· 4分 （2）解法一： 连结OA OB ，和OC ，则AOC COB BOA △≌△≌△，12∠=∠， ·
·························· 5分 不妨设OD 交BC 于点F ，OE 交AC 于点G ，
3412054120AOC DOE ∠=∠+∠=∠=∠+∠=°，°，
35∴∠=∠． ······································································· 7分 在OAG △和OCF △中， 1235OA OC ∠=∠⎧⎪
=⎨⎪∠=∠⎩
，，，
OAG OCF ∴△≌△， ························································································· 8分 1
3OFCG AOC ABC S S S ∴==△△． ·
·············································································· 9分 解法二： 不妨设OD 交BC 于点F ，OE 交AC 于点G ， 作OH BC OK AC ⊥⊥，，垂足分别为H K 、， ·················· 5分 在四边形HOKC 中，9060OHC OKC C ∠=∠=∠=°，°， 360909060120HOK ∴∠=-︒-︒=︒°-?， ·
······················· 6分 即12120∠+∠=°．
又23120GOF ∠=∠+∠=°，
13∴∠=∠． ·
···································································································· 7分 AC BC =， OH OK ∴=，
OGK OFH ∴△≌△， ·
······················································································· 8分 1
3
OFCG OHCK ABC S S S ∴==△． ·
··············································································· 9分
答案20题图（2）
A E O G
F
B C D 1 2 3 4
5 答案第20题图（3） A E
O
G
F B C D 1 3 2
H K。